# Prove that for density operators, if $\rho=\sum_i \rho_i P_i$ then $\log\rho=\sum_i P_i\log \rho_i$

I want to prove what's used in the fourth line below the "Proof" section here: http://en.wikipedia.org/wiki/Quantum_relative_entropy#The_result

The statement is: Let $\rho$ be a density operator on a finite dimensional complex inner product space, with spectral decomposition $\rho = \sum_i \rho_i P_i$, with orthogonal projections $P_i$ and eigenvalues $\rho_i$. Then $$\log\rho = \sum_i P_i\log\rho_i$$

I don't see how to prove this. Of course, if I imagine the projections to be zero-matrices with a $1$ somewhere on the diagonal such that they all add up to $\mathbb{1}$, then it makes perfect sense, as then I just have the logarithm of a diagonal matrix, which is clear. But how can I see this rigorously from the properties of orthogonal projections?

• Wikipedia's page on matrix logarithms may be helpful. Also note that it's sufficient to check that $\exp \log \rho$ is equal to $\rho$. Jul 21, 2014 at 15:36
• That was actually quite helpful! Jul 21, 2014 at 15:55
• Glad to hear it. Keep in mind that you're allowed/encouraged to answer your own questions, and (after a sufficient time period) will be able to accept it as the answer if you consider that the best answer. That also lets us give feedback etc in comments. Jul 21, 2014 at 15:56

Using $P_i^2 = P_i$ we get $$e^{P_i\log\rho_i} = \sum_{k=0}^\infty \frac{(P_i\log\rho_i)^k}{k!} = \mathbb{1}-P_i +P_i\sum\frac{\log^k\rho_i}{k!} = \mathbb{1}-P_i+P_i\rho_i$$ For the whole expression we find (using that $P_iP_k=\delta_{i,k}P_k$, and $\sum_iP_i=\mathbb{1}$) $$\exp \sum_iP_i\log\rho_i = \prod_i\mathbb{1}-P_i+P_i\rho_i = \mathbb{1} -\sum_iP_i + \rho = \rho.$$ Hence indeed $\log\rho = \sum_iP_i\log\rho_i$.